Posted by Hannah on Monday, March 7, 2011 at 3:14pm.
I posted this question a few days ago and someone answered it but I just wanted to make sure that I copied down the correct answers and didn't misunderstand anything.
Gas is escaping a spherical balloon at the rate of 2ft^3/min. How fast is the surface area shrinking (ds/dt) when the radius is 12ft? ( A sphere of raius r has volume= 4/3 pir^3 and surface area S=4pir^2.)
Remember that ds/dt=ds/dr X dr/dt
Step 1: Find ds/dr
S=4pir^2 so ds/dr= (4pi)(2r) = 8pi r
Step 2: Fid dr/dt (Hint: dv/dt=dv/dr X dr/dt)
V=4/3pir^3 dv/dr=4pir^2 2=(4pir^2)(dr/dt) dr/dt=1/2pir^2
Step 3: Find ds/dt
ds/dt= (8pir)(1/2pir^2) = 4/r
Step 4: Evaluate ds/dt when the radius is 12ft.
r=12 (8)(3.14)(12) X (1/2(3.14)(12)^2) =4/12
Are these correct?
- Math(Please check) - Reiny, Monday, March 7, 2011 at 3:41pm
correct, except reduce 4/12 to 1/3
I might have done it this way
V = (4/3)πr^3
dV/dt = 4πr^2 dr/dt , but when r=12, dV/dt = 2
2 = 4π(144) dr/dt
dr/dt = 1/(288π)
A = 4πr^2
dA/dt = 8πr dr/dt
= 8π(12)(1/(288π) = 1/3
- Math(Please check) - Hannah, Monday, March 7, 2011 at 3:45pm
Ok thank you!
- Math(Please check) - Reiny, Monday, March 7, 2011 at 4:09pm
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